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$Let S_{n} denote the sum of the first n-terms of an A.P. If S_{2 n} = 3 \cdot S_{n} then S_{3 n} \div S_{n}$

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Correct option is B

S2n=3.Sn⇒2n2[2a+(2n−1)d]=3⋅n2[2a+(n−1)d]⇒4a+22n-1d=6a+3n-1d⇒2a=(n+1)d∴s3nsn=3n22a+(3n−1)dn[2a+(n−1)d2=3(4nd)2nd=6

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Let Sn denote the sum of the first n-terms of an A.P. If S2n=3⋅Sn then S3n÷Sn

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$Let S_{n} denote the sum of the first n-terms of an A.P. If S_{2 n} = 3 \cdot S_{n} then S_{3 n} \div S_{n}$